Mixing
An extension $K/F$ is Galois if for every simple extension $F subset F(u) subset K$, $[F(u):F] leq 2$.
$begingroup$
Let $F$ be a field, $char(F)neq 2$ and let $K$ be an extension field of $F$. If for each $uin K$, $[F(u):F]leq 2$, show that $K$ is a Galois extension of $F$.
field-theory galois-theory extension-field galois-extensions
edited Jan 24 at 4:50
Brahadeesh
6,50942363
asked Jan 23 at 17:35
BabaiBabai
2,65121640
$endgroup$
add a comment |
$begingroup$
Let $F$ be a field, $char(F)neq 2$ and let $K$ be an extension field of $F$. If for each $uin K$, $[F(u):F]leq 2$, show that $K$ is a Galois extension of $F$.
field-theory galois-theory extension-field galois-extensions
edited Jan 24 at 4:50
Brahadeesh
6,50942363
asked Jan 23 at 17:35
BabaiBabai
2,65121640
$endgroup$
1
$begingroup$
Then every element is separable over $F$ and the primitive element theorem gives $K = F(u)$. The minimal polynomial then shows $K/F$ is Galois. Also your title should be "if every simple extension contained in it"
$endgroup$
– reuns
Jan 23 at 19:30
add a comment |
$begingroup$
Let $F$ be a field, $char(F)neq 2$ and let $K$ be an extension field of $F$. If for each $uin K$, $[F(u):F]leq 2$, show that $K$ is a Galois extension of $F$.
field-theory galois-theory extension-field galois-extensions
edited Jan 24 at 4:50
Brahadeesh
6,50942363
asked Jan 23 at 17:35
BabaiBabai
2,65121640
$endgroup$
Let $F$ be a field, $char(F)neq 2$ and let $K$ be an extension field of $F$. If for each $uin K$, $[F(u):F]leq 2$, show that $K$ is a Galois extension of $F$.
field-theory galois-theory extension-field galois-extensions
field-theory galois-theory extension-field galois-extensions
edited Jan 24 at 4:50
Brahadeesh
6,50942363
asked Jan 23 at 17:35
BabaiBabai
2,65121640
edited Jan 24 at 4:50
Brahadeesh
6,50942363
asked Jan 23 at 17:35
BabaiBabai
2,65121640
edited Jan 24 at 4:50
Brahadeesh
6,50942363
edited Jan 24 at 4:50
Brahadeesh
6,50942363
edited Jan 24 at 4:50
Brahadeesh
6,50942363
6,50942363
asked Jan 23 at 17:35
BabaiBabai
2,65121640
asked Jan 23 at 17:35
BabaiBabai
2,65121640
asked Jan 23 at 17:35
BabaiBabai
2,65121640
2,65121640
1
$begingroup$
Then every element is separable over $F$ and the primitive element theorem gives $K = F(u)$. The minimal polynomial then shows $K/F$ is Galois. Also your title should be "if every simple extension contained in it"
$endgroup$
– reuns
Jan 23 at 19:30
add a comment |
1
$begingroup$
Then every element is separable over $F$ and the primitive element theorem gives $K = F(u)$. The minimal polynomial then shows $K/F$ is Galois. Also your title should be "if every simple extension contained in it"
$endgroup$
– reuns
Jan 23 at 19:30
1
1
$begingroup$
Then every element is separable over $F$ and the primitive element theorem gives $K = F(u)$. The minimal polynomial then shows $K/F$ is Galois. Also your title should be "if every simple extension contained in it"
$endgroup$
– reuns
Jan 23 at 19:30
$begingroup$
Then every element is separable over $F$ and the primitive element theorem gives $K = F(u)$. The minimal polynomial then shows $K/F$ is Galois. Also your title should be "if every simple extension contained in it"
$endgroup$
– reuns
Jan 23 at 19:30
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Here's a roadmap:
An extension of fields $K/F$ is Galois if it is normal and separable.
Every degree $2$ extension is normal and a compositum of normal extensions is normal.
An element $u in K$ is separable over $F$ if its minimal polynomial has distinct roots. A polynomial $f$ has distinct roots if and only if $gcd(f,f')=1$.
answered Jan 24 at 4:47
BrahadeeshBrahadeesh
6,50942363
$endgroup$
$begingroup$
Are you sure "compositum of normal extensions is normal" is true? I mean $mathbb{Q} subseteq mathbb{Q}(sqrt{2}) subseteq mathbb{Q}(sqrt[4]{2})$ is a counterexample. The two intermediate extensions are normal, but their composition $mathbb{Q} subset mathbb{Q}(sqrt[4]{2})$ isn't. Or maybe I have misunderstood something?
$endgroup$
– Stefan4024
Jan 25 at 12:20
2
$begingroup$
@Stefan4024 what I meant is that if $k subseteq F_i subseteq K$, $i in I$, with each $F_i$ normal over $k$, then the compositum of the $F_i$’s is also normal over $k$.
$endgroup$
– Brahadeesh
Jan 25 at 14:47
add a comment |
$begingroup$
We need to show that $K$ is both separable and normal over $F$, for that is what it means to be Galois.
We are given that
$u in K Longrightarrow [F(u):F] le 2; tag 1$
let
$m(x) in F[x] tag 2$
be the minimal polynomial of $u in K$ over $F$; then in accord with (1) we have
$deg m(x) le 2; tag 3$
now,
$deg m(x) = 1 Longleftrightarrow m(x) = x + b, ; b in F, tag 4$
so
$m(u) = u + b = 0 Longleftrightarrow u = -b in F; tag 5$
we thus see that when $[F(u):F] = 1$, the minimal polynomial $m(x) = x + b$ is manifestly separable, having as it does exactly one root $-b$, which obviously is not repeated. As for the case $[F(u):F] = 2$, we may write
$m(x) = x^2 + ax + b; tag 6$
here the possibility that
$b = dfrac{a^2}{4} tag 7$
is precluded by the assumption $[F(u):F] = 2$, lest
$m(x) = x^2 + ax + dfrac{a^2}{4} = left ( x + dfrac{a}{2} right )^2, tag 8$
which has a double root $u = -frac{a}{2} in F$ but then $F(u) = F$ so that again $[F(u):F] = 1$, a contradiction; thus
$b ne dfrac{a^2}{4}; tag 9$
from (6) we find that
$m'(x) = 2x + a; tag{10}$
which has one root
$u = -dfrac{a}{2}; tag{11}$
is this root shared by $m(x)$? if so, then
$m left (-dfrac{a}{2} right ) = -dfrac{a^2}{4} + b = 0; tag{12}$
which holds if (7) does also; so in the light of (9) we find
$m left (-dfrac{a}{2} right ) = -dfrac{a^2}{4} + b ne 0, tag{13}$
and conclude that since $m(x)$ and $m'(x)$ have no shared roots, $m(x)$ is a separable quadratic and therefore the field extension $F(u)/F$ is separable is well. Hence the extension $K/F$ is also separable.
Having established that $K/F$ is separable, we now turn to the question of normalcy.
Let
$p(x) in F[x] tag{14}$
be irreducible, and suppose $p(x)$ has a root
$r in K; tag{15}$
that is,
$p(r) = 0; tag{16}$
we may assume, without loss of generality, that $p(x)$ is monic; by hypothesis,
$[F(r):F] le 2; tag{17}$
we may rule out the possibility that $[F(r):F] = 1$, since then $r in F$ and then $x - r mid p(x)$ in $F$, contrary to the assumed irreducibility of $p(x)$; so in fact
$[F(r):F] = 2; tag{18}$
it follows then that we may take the minimal polynomial of $r$ over $F$ to be a quadratic $m(x)$ of the form (6):
$m(x) = x^2 + ax + b in F[x], ; a, b in F tag{19}$
with
$m(r) = 0; tag{20}$
I claim
$m(-(a + r)) = 0; tag{21}$
for
$m(-(a + r)) = (a + r)^2 - a(a + r) + b$
$= a^2 + 2ar + r^2 - a^2 - ar + b = r^2 + ar + b = 0; tag{22}$
thus $-(a + r)$ is also a root of $m(x)$ which is then seen to split in $K$
$m(x) = (x - r)(x - (r + a)); tag{23}$
furthermore, since $m(x)$ is minimal for $r$ over $F$, we have in light of (14) that
$m(x) mid p(x); tag{24}$
now since $p(x)$ is irreducible we may infer from (24) that
$m(x) = p(x); tag{25}$
hence $p(x)$ splits in $K$ whence $K$ is normal over $F$.
Since $K$ is both normal and separable over $F$, by definition $K$ is a Galois extension of the field $F$.
answered Jan 28 at 2:12
Robert LewisRobert Lewis
48.1k23067
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
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active
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oldest
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$begingroup$
Here's a roadmap:
An extension of fields $K/F$ is Galois if it is normal and separable.
Every degree $2$ extension is normal and a compositum of normal extensions is normal.
An element $u in K$ is separable over $F$ if its minimal polynomial has distinct roots. A polynomial $f$ has distinct roots if and only if $gcd(f,f')=1$.
answered Jan 24 at 4:47
BrahadeeshBrahadeesh
6,50942363
$endgroup$
$begingroup$
Are you sure "compositum of normal extensions is normal" is true? I mean $mathbb{Q} subseteq mathbb{Q}(sqrt{2}) subseteq mathbb{Q}(sqrt[4]{2})$ is a counterexample. The two intermediate extensions are normal, but their composition $mathbb{Q} subset mathbb{Q}(sqrt[4]{2})$ isn't. Or maybe I have misunderstood something?
$endgroup$
– Stefan4024
Jan 25 at 12:20
2
$begingroup$
@Stefan4024 what I meant is that if $k subseteq F_i subseteq K$, $i in I$, with each $F_i$ normal over $k$, then the compositum of the $F_i$’s is also normal over $k$.
$endgroup$
– Brahadeesh
Jan 25 at 14:47
add a comment |
$begingroup$
Here's a roadmap:
An extension of fields $K/F$ is Galois if it is normal and separable.
Every degree $2$ extension is normal and a compositum of normal extensions is normal.
An element $u in K$ is separable over $F$ if its minimal polynomial has distinct roots. A polynomial $f$ has distinct roots if and only if $gcd(f,f')=1$.
answered Jan 24 at 4:47
BrahadeeshBrahadeesh
6,50942363
$endgroup$
$begingroup$
Are you sure "compositum of normal extensions is normal" is true? I mean $mathbb{Q} subseteq mathbb{Q}(sqrt{2}) subseteq mathbb{Q}(sqrt[4]{2})$ is a counterexample. The two intermediate extensions are normal, but their composition $mathbb{Q} subset mathbb{Q}(sqrt[4]{2})$ isn't. Or maybe I have misunderstood something?
$endgroup$
– Stefan4024
Jan 25 at 12:20
2
$begingroup$
@Stefan4024 what I meant is that if $k subseteq F_i subseteq K$, $i in I$, with each $F_i$ normal over $k$, then the compositum of the $F_i$’s is also normal over $k$.
$endgroup$
– Brahadeesh
Jan 25 at 14:47
add a comment |
$begingroup$
Here's a roadmap:
An extension of fields $K/F$ is Galois if it is normal and separable.
Every degree $2$ extension is normal and a compositum of normal extensions is normal.
An element $u in K$ is separable over $F$ if its minimal polynomial has distinct roots. A polynomial $f$ has distinct roots if and only if $gcd(f,f')=1$.
answered Jan 24 at 4:47
BrahadeeshBrahadeesh
6,50942363
$endgroup$
Here's a roadmap:
An extension of fields $K/F$ is Galois if it is normal and separable.
Every degree $2$ extension is normal and a compositum of normal extensions is normal.
An element $u in K$ is separable over $F$ if its minimal polynomial has distinct roots. A polynomial $f$ has distinct roots if and only if $gcd(f,f')=1$.
answered Jan 24 at 4:47
BrahadeeshBrahadeesh
6,50942363
answered Jan 24 at 4:47
BrahadeeshBrahadeesh
6,50942363
answered Jan 24 at 4:47
BrahadeeshBrahadeesh
6,50942363
answered Jan 24 at 4:47
BrahadeeshBrahadeesh
6,50942363
6,50942363
$begingroup$
Are you sure "compositum of normal extensions is normal" is true? I mean $mathbb{Q} subseteq mathbb{Q}(sqrt{2}) subseteq mathbb{Q}(sqrt[4]{2})$ is a counterexample. The two intermediate extensions are normal, but their composition $mathbb{Q} subset mathbb{Q}(sqrt[4]{2})$ isn't. Or maybe I have misunderstood something?
$endgroup$
– Stefan4024
Jan 25 at 12:20
2
$begingroup$
@Stefan4024 what I meant is that if $k subseteq F_i subseteq K$, $i in I$, with each $F_i$ normal over $k$, then the compositum of the $F_i$’s is also normal over $k$.
$endgroup$
– Brahadeesh
Jan 25 at 14:47
add a comment |
$begingroup$
Are you sure "compositum of normal extensions is normal" is true? I mean $mathbb{Q} subseteq mathbb{Q}(sqrt{2}) subseteq mathbb{Q}(sqrt[4]{2})$ is a counterexample. The two intermediate extensions are normal, but their composition $mathbb{Q} subset mathbb{Q}(sqrt[4]{2})$ isn't. Or maybe I have misunderstood something?
$endgroup$
– Stefan4024
Jan 25 at 12:20
2
$begingroup$
@Stefan4024 what I meant is that if $k subseteq F_i subseteq K$, $i in I$, with each $F_i$ normal over $k$, then the compositum of the $F_i$’s is also normal over $k$.
$endgroup$
– Brahadeesh
Jan 25 at 14:47
$begingroup$
Are you sure "compositum of normal extensions is normal" is true? I mean $mathbb{Q} subseteq mathbb{Q}(sqrt{2}) subseteq mathbb{Q}(sqrt[4]{2})$ is a counterexample. The two intermediate extensions are normal, but their composition $mathbb{Q} subset mathbb{Q}(sqrt[4]{2})$ isn't. Or maybe I have misunderstood something?
$endgroup$
– Stefan4024
Jan 25 at 12:20
$begingroup$
Are you sure "compositum of normal extensions is normal" is true? I mean $mathbb{Q} subseteq mathbb{Q}(sqrt{2}) subseteq mathbb{Q}(sqrt[4]{2})$ is a counterexample. The two intermediate extensions are normal, but their composition $mathbb{Q} subset mathbb{Q}(sqrt[4]{2})$ isn't. Or maybe I have misunderstood something?
$endgroup$
– Stefan4024
Jan 25 at 12:20
2
2
$begingroup$
@Stefan4024 what I meant is that if $k subseteq F_i subseteq K$, $i in I$, with each $F_i$ normal over $k$, then the compositum of the $F_i$’s is also normal over $k$.
$endgroup$
– Brahadeesh
Jan 25 at 14:47
$begingroup$
@Stefan4024 what I meant is that if $k subseteq F_i subseteq K$, $i in I$, with each $F_i$ normal over $k$, then the compositum of the $F_i$’s is also normal over $k$.
$endgroup$
– Brahadeesh
Jan 25 at 14:47
add a comment |
$begingroup$
We need to show that $K$ is both separable and normal over $F$, for that is what it means to be Galois.
We are given that
$u in K Longrightarrow [F(u):F] le 2; tag 1$
let
$m(x) in F[x] tag 2$
be the minimal polynomial of $u in K$ over $F$; then in accord with (1) we have
$deg m(x) le 2; tag 3$
now,
$deg m(x) = 1 Longleftrightarrow m(x) = x + b, ; b in F, tag 4$
so
$m(u) = u + b = 0 Longleftrightarrow u = -b in F; tag 5$
we thus see that when $[F(u):F] = 1$, the minimal polynomial $m(x) = x + b$ is manifestly separable, having as it does exactly one root $-b$, which obviously is not repeated. As for the case $[F(u):F] = 2$, we may write
$m(x) = x^2 + ax + b; tag 6$
here the possibility that
$b = dfrac{a^2}{4} tag 7$
is precluded by the assumption $[F(u):F] = 2$, lest
$m(x) = x^2 + ax + dfrac{a^2}{4} = left ( x + dfrac{a}{2} right )^2, tag 8$
which has a double root $u = -frac{a}{2} in F$ but then $F(u) = F$ so that again $[F(u):F] = 1$, a contradiction; thus
$b ne dfrac{a^2}{4}; tag 9$
from (6) we find that
$m'(x) = 2x + a; tag{10}$
which has one root
$u = -dfrac{a}{2}; tag{11}$
is this root shared by $m(x)$? if so, then
$m left (-dfrac{a}{2} right ) = -dfrac{a^2}{4} + b = 0; tag{12}$
which holds if (7) does also; so in the light of (9) we find
$m left (-dfrac{a}{2} right ) = -dfrac{a^2}{4} + b ne 0, tag{13}$
and conclude that since $m(x)$ and $m'(x)$ have no shared roots, $m(x)$ is a separable quadratic and therefore the field extension $F(u)/F$ is separable is well. Hence the extension $K/F$ is also separable.
Having established that $K/F$ is separable, we now turn to the question of normalcy.
Let
$p(x) in F[x] tag{14}$
be irreducible, and suppose $p(x)$ has a root
$r in K; tag{15}$
that is,
$p(r) = 0; tag{16}$
we may assume, without loss of generality, that $p(x)$ is monic; by hypothesis,
$[F(r):F] le 2; tag{17}$
we may rule out the possibility that $[F(r):F] = 1$, since then $r in F$ and then $x - r mid p(x)$ in $F$, contrary to the assumed irreducibility of $p(x)$; so in fact
$[F(r):F] = 2; tag{18}$
it follows then that we may take the minimal polynomial of $r$ over $F$ to be a quadratic $m(x)$ of the form (6):
$m(x) = x^2 + ax + b in F[x], ; a, b in F tag{19}$
with
$m(r) = 0; tag{20}$
I claim
$m(-(a + r)) = 0; tag{21}$
for
$m(-(a + r)) = (a + r)^2 - a(a + r) + b$
$= a^2 + 2ar + r^2 - a^2 - ar + b = r^2 + ar + b = 0; tag{22}$
thus $-(a + r)$ is also a root of $m(x)$ which is then seen to split in $K$
$m(x) = (x - r)(x - (r + a)); tag{23}$
furthermore, since $m(x)$ is minimal for $r$ over $F$, we have in light of (14) that
$m(x) mid p(x); tag{24}$
now since $p(x)$ is irreducible we may infer from (24) that
$m(x) = p(x); tag{25}$
hence $p(x)$ splits in $K$ whence $K$ is normal over $F$.
Since $K$ is both normal and separable over $F$, by definition $K$ is a Galois extension of the field $F$.
answered Jan 28 at 2:12
Robert LewisRobert Lewis
48.1k23067
$endgroup$
add a comment |
$begingroup$
We need to show that $K$ is both separable and normal over $F$, for that is what it means to be Galois.
We are given that
$u in K Longrightarrow [F(u):F] le 2; tag 1$
let
$m(x) in F[x] tag 2$
be the minimal polynomial of $u in K$ over $F$; then in accord with (1) we have
$deg m(x) le 2; tag 3$
now,
$deg m(x) = 1 Longleftrightarrow m(x) = x + b, ; b in F, tag 4$
so
$m(u) = u + b = 0 Longleftrightarrow u = -b in F; tag 5$
we thus see that when $[F(u):F] = 1$, the minimal polynomial $m(x) = x + b$ is manifestly separable, having as it does exactly one root $-b$, which obviously is not repeated. As for the case $[F(u):F] = 2$, we may write
$m(x) = x^2 + ax + b; tag 6$
here the possibility that
$b = dfrac{a^2}{4} tag 7$
is precluded by the assumption $[F(u):F] = 2$, lest
$m(x) = x^2 + ax + dfrac{a^2}{4} = left ( x + dfrac{a}{2} right )^2, tag 8$
which has a double root $u = -frac{a}{2} in F$ but then $F(u) = F$ so that again $[F(u):F] = 1$, a contradiction; thus
$b ne dfrac{a^2}{4}; tag 9$
from (6) we find that
$m'(x) = 2x + a; tag{10}$
which has one root
$u = -dfrac{a}{2}; tag{11}$
is this root shared by $m(x)$? if so, then
$m left (-dfrac{a}{2} right ) = -dfrac{a^2}{4} + b = 0; tag{12}$
which holds if (7) does also; so in the light of (9) we find
$m left (-dfrac{a}{2} right ) = -dfrac{a^2}{4} + b ne 0, tag{13}$
and conclude that since $m(x)$ and $m'(x)$ have no shared roots, $m(x)$ is a separable quadratic and therefore the field extension $F(u)/F$ is separable is well. Hence the extension $K/F$ is also separable.
Having established that $K/F$ is separable, we now turn to the question of normalcy.
Let
$p(x) in F[x] tag{14}$
be irreducible, and suppose $p(x)$ has a root
$r in K; tag{15}$
that is,
$p(r) = 0; tag{16}$
we may assume, without loss of generality, that $p(x)$ is monic; by hypothesis,
$[F(r):F] le 2; tag{17}$
we may rule out the possibility that $[F(r):F] = 1$, since then $r in F$ and then $x - r mid p(x)$ in $F$, contrary to the assumed irreducibility of $p(x)$; so in fact
$[F(r):F] = 2; tag{18}$
it follows then that we may take the minimal polynomial of $r$ over $F$ to be a quadratic $m(x)$ of the form (6):
$m(x) = x^2 + ax + b in F[x], ; a, b in F tag{19}$
with
$m(r) = 0; tag{20}$
I claim
$m(-(a + r)) = 0; tag{21}$
for
$m(-(a + r)) = (a + r)^2 - a(a + r) + b$
$= a^2 + 2ar + r^2 - a^2 - ar + b = r^2 + ar + b = 0; tag{22}$
thus $-(a + r)$ is also a root of $m(x)$ which is then seen to split in $K$
$m(x) = (x - r)(x - (r + a)); tag{23}$
furthermore, since $m(x)$ is minimal for $r$ over $F$, we have in light of (14) that
$m(x) mid p(x); tag{24}$
now since $p(x)$ is irreducible we may infer from (24) that
$m(x) = p(x); tag{25}$
hence $p(x)$ splits in $K$ whence $K$ is normal over $F$.
Since $K$ is both normal and separable over $F$, by definition $K$ is a Galois extension of the field $F$.
answered Jan 28 at 2:12
Robert LewisRobert Lewis
48.1k23067
$endgroup$
add a comment |
$begingroup$
We need to show that $K$ is both separable and normal over $F$, for that is what it means to be Galois.
We are given that
$u in K Longrightarrow [F(u):F] le 2; tag 1$
let
$m(x) in F[x] tag 2$
be the minimal polynomial of $u in K$ over $F$; then in accord with (1) we have
$deg m(x) le 2; tag 3$
now,
$deg m(x) = 1 Longleftrightarrow m(x) = x + b, ; b in F, tag 4$
so
$m(u) = u + b = 0 Longleftrightarrow u = -b in F; tag 5$
we thus see that when $[F(u):F] = 1$, the minimal polynomial $m(x) = x + b$ is manifestly separable, having as it does exactly one root $-b$, which obviously is not repeated. As for the case $[F(u):F] = 2$, we may write
$m(x) = x^2 + ax + b; tag 6$
here the possibility that
$b = dfrac{a^2}{4} tag 7$
is precluded by the assumption $[F(u):F] = 2$, lest
$m(x) = x^2 + ax + dfrac{a^2}{4} = left ( x + dfrac{a}{2} right )^2, tag 8$
which has a double root $u = -frac{a}{2} in F$ but then $F(u) = F$ so that again $[F(u):F] = 1$, a contradiction; thus
$b ne dfrac{a^2}{4}; tag 9$
from (6) we find that
$m'(x) = 2x + a; tag{10}$
which has one root
$u = -dfrac{a}{2}; tag{11}$
is this root shared by $m(x)$? if so, then
$m left (-dfrac{a}{2} right ) = -dfrac{a^2}{4} + b = 0; tag{12}$
which holds if (7) does also; so in the light of (9) we find
$m left (-dfrac{a}{2} right ) = -dfrac{a^2}{4} + b ne 0, tag{13}$
and conclude that since $m(x)$ and $m'(x)$ have no shared roots, $m(x)$ is a separable quadratic and therefore the field extension $F(u)/F$ is separable is well. Hence the extension $K/F$ is also separable.
Having established that $K/F$ is separable, we now turn to the question of normalcy.
Let
$p(x) in F[x] tag{14}$
be irreducible, and suppose $p(x)$ has a root
$r in K; tag{15}$
that is,
$p(r) = 0; tag{16}$
we may assume, without loss of generality, that $p(x)$ is monic; by hypothesis,
$[F(r):F] le 2; tag{17}$
we may rule out the possibility that $[F(r):F] = 1$, since then $r in F$ and then $x - r mid p(x)$ in $F$, contrary to the assumed irreducibility of $p(x)$; so in fact
$[F(r):F] = 2; tag{18}$
it follows then that we may take the minimal polynomial of $r$ over $F$ to be a quadratic $m(x)$ of the form (6):
$m(x) = x^2 + ax + b in F[x], ; a, b in F tag{19}$
with
$m(r) = 0; tag{20}$
I claim
$m(-(a + r)) = 0; tag{21}$
for
$m(-(a + r)) = (a + r)^2 - a(a + r) + b$
$= a^2 + 2ar + r^2 - a^2 - ar + b = r^2 + ar + b = 0; tag{22}$
thus $-(a + r)$ is also a root of $m(x)$ which is then seen to split in $K$
$m(x) = (x - r)(x - (r + a)); tag{23}$
furthermore, since $m(x)$ is minimal for $r$ over $F$, we have in light of (14) that
$m(x) mid p(x); tag{24}$
now since $p(x)$ is irreducible we may infer from (24) that
$m(x) = p(x); tag{25}$
hence $p(x)$ splits in $K$ whence $K$ is normal over $F$.
Since $K$ is both normal and separable over $F$, by definition $K$ is a Galois extension of the field $F$.
answered Jan 28 at 2:12
Robert LewisRobert Lewis
48.1k23067
$endgroup$
We need to show that $K$ is both separable and normal over $F$, for that is what it means to be Galois.
We are given that
$u in K Longrightarrow [F(u):F] le 2; tag 1$
let
$m(x) in F[x] tag 2$
be the minimal polynomial of $u in K$ over $F$; then in accord with (1) we have
$deg m(x) le 2; tag 3$
now,
$deg m(x) = 1 Longleftrightarrow m(x) = x + b, ; b in F, tag 4$
so
$m(u) = u + b = 0 Longleftrightarrow u = -b in F; tag 5$
we thus see that when $[F(u):F] = 1$, the minimal polynomial $m(x) = x + b$ is manifestly separable, having as it does exactly one root $-b$, which obviously is not repeated. As for the case $[F(u):F] = 2$, we may write
$m(x) = x^2 + ax + b; tag 6$
here the possibility that
$b = dfrac{a^2}{4} tag 7$
is precluded by the assumption $[F(u):F] = 2$, lest
$m(x) = x^2 + ax + dfrac{a^2}{4} = left ( x + dfrac{a}{2} right )^2, tag 8$
which has a double root $u = -frac{a}{2} in F$ but then $F(u) = F$ so that again $[F(u):F] = 1$, a contradiction; thus
$b ne dfrac{a^2}{4}; tag 9$
from (6) we find that
$m'(x) = 2x + a; tag{10}$
which has one root
$u = -dfrac{a}{2}; tag{11}$
is this root shared by $m(x)$? if so, then
$m left (-dfrac{a}{2} right ) = -dfrac{a^2}{4} + b = 0; tag{12}$
which holds if (7) does also; so in the light of (9) we find
$m left (-dfrac{a}{2} right ) = -dfrac{a^2}{4} + b ne 0, tag{13}$
and conclude that since $m(x)$ and $m'(x)$ have no shared roots, $m(x)$ is a separable quadratic and therefore the field extension $F(u)/F$ is separable is well. Hence the extension $K/F$ is also separable.
Having established that $K/F$ is separable, we now turn to the question of normalcy.
Let
$p(x) in F[x] tag{14}$
be irreducible, and suppose $p(x)$ has a root
$r in K; tag{15}$
that is,
$p(r) = 0; tag{16}$
we may assume, without loss of generality, that $p(x)$ is monic; by hypothesis,
$[F(r):F] le 2; tag{17}$
we may rule out the possibility that $[F(r):F] = 1$, since then $r in F$ and then $x - r mid p(x)$ in $F$, contrary to the assumed irreducibility of $p(x)$; so in fact
$[F(r):F] = 2; tag{18}$
it follows then that we may take the minimal polynomial of $r$ over $F$ to be a quadratic $m(x)$ of the form (6):
$m(x) = x^2 + ax + b in F[x], ; a, b in F tag{19}$
with
$m(r) = 0; tag{20}$
I claim
$m(-(a + r)) = 0; tag{21}$
for
$m(-(a + r)) = (a + r)^2 - a(a + r) + b$
$= a^2 + 2ar + r^2 - a^2 - ar + b = r^2 + ar + b = 0; tag{22}$
thus $-(a + r)$ is also a root of $m(x)$ which is then seen to split in $K$
$m(x) = (x - r)(x - (r + a)); tag{23}$
furthermore, since $m(x)$ is minimal for $r$ over $F$, we have in light of (14) that
$m(x) mid p(x); tag{24}$
now since $p(x)$ is irreducible we may infer from (24) that
$m(x) = p(x); tag{25}$
hence $p(x)$ splits in $K$ whence $K$ is normal over $F$.
Since $K$ is both normal and separable over $F$, by definition $K$ is a Galois extension of the field $F$.
answered Jan 28 at 2:12
Robert LewisRobert Lewis
48.1k23067
edited Jan 28 at 2:36
answered Jan 28 at 2:12
Robert LewisRobert Lewis
48.1k23067
answered Jan 28 at 2:12
Robert LewisRobert Lewis
48.1k23067
answered Jan 28 at 2:12
Robert LewisRobert Lewis
48.1k23067
48.1k23067
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$begingroup$
Then every element is separable over $F$ and the primitive element theorem gives $K = F(u)$. The minimal polynomial then shows $K/F$ is Galois. Also your title should be "if every simple extension contained in it"
$endgroup$
– reuns
Jan 23 at 19:30